Proof of Bounded Lower Semicontinuous Functions are Increasing Limits of Lipschitz Functions

Throughout, write $S_{k,x}=\{f(y)+\lambda_k\,d(x,y):y\in X\}$ for $k\in\mathbb{N}$ and $x\in X$, and let $d_{\mathbb{R}}$ be the absolute-value metric on $\mathbb{R}$. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $0<\lambda_k$ for every $k\in\mathbb{N}$.

Claim 1. Since $X$ is nonempty, $S_{k,x}$ is nonempty. Every element of $S_{k,x}$ is nonnegative, because $0\le f(y)$ by hypothesis and $0\le\lambda_k\,d(x,y)$ by condition 1 of Metric Space together with $0<\lambda_k$; so $0$ is a lower bound of $S_{k,x}$. Hence the infimum of $S_{k,x}$ exists by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below, is unique by Uniqueness of the Supremum and of the Infimum, and satisfies $0\le f_k(x)$. Taking $y=x$ and using $d(x,x)=0$ (condition 2 of Metric Space) shows that $f(x)\in S_{k,x}$, so $f_k(x)\le f(x)$; and $f(x)\le M$ by hypothesis.

Claim 2. Let $x,x'\in X$ and let $y\in X$ be arbitrary. Since $f_k(x)$ is a lower bound of $S_{k,x}$, and by the triangle inequality (condition 4 of Metric Space) together with $0\le\lambda_k$,

Hence $f_k(x)-\lambda_k\,d(x,x')\le f(y)+\lambda_k\,d(x',y)$ for every $y\in X$, so $f_k(x)-\lambda_k\,d(x,x')$ is a lower bound of $S_{k,x'}$ and therefore

Interchanging $x$ and $x'$ and using the symmetry of $d$ (condition 3 of Metric Space) gives $f_k(x')-f_k(x)\le\lambda_k\,d(x,x')$. By claim 6 of Properties of the Absolute Value in an Ordered Field, the two inequalities give

so $f_k$ is Lipschitz with constant $\lambda_k$, and hence continuous on $X$ by A Lipschitz Map is Uniformly Continuous.

Claim 3. By claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, $\lambda_k<\lambda_{k+1}$. For every $y\in X$ we have $0\le d(x,y)$, so $\lambda_k\,d(x,y)\le\lambda_{k+1}\,d(x,y)$ and therefore

Thus $f_k(x)$ is a lower bound of $S_{k+1,x}$, whence $f_k(x)\le f_{k+1}(x)$.

Claim 4. Fix $x\in X$ and let $\varepsilon$ be a real number with $0<\varepsilon$. Since $f$ is lower semicontinuous at $x$ relative to $X$, there is a real $\delta>0$ such that every $y\in X$ with $d(x,y)<\delta$ satisfies

By claim 2 of The Archimedean Property of the Real Numbers there is $K\in\mathbb{N}$ with $M<\lambda_K\,\delta$. Let $k\in\mathbb{N}$ with $K\le k$; then $\lambda_K\le\lambda_k$, by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field if $K<k$ and trivially if $K=k$.

Let $y\in X$. If $d(x,y)<\delta$, then $0\le\lambda_k\,d(x,y)$ gives $f(x)-\tfrac{\varepsilon}{2}<f(y)\le f(y)+\lambda_k\,d(x,y)$. If instead $\delta\le d(x,y)$, then

using $0\le f(y)$ and $0<\varepsilon$. In both cases $f(x)-\tfrac{\varepsilon}{2}<f(y)+\lambda_k\,d(x,y)$, so $f(x)-\tfrac{\varepsilon}{2}$ is a lower bound of $S_{k,x}$ and therefore $f(x)-\tfrac{\varepsilon}{2}\le f_k(x)$.

Combining with claim 1, $f(x)-\tfrac{\varepsilon}{2}\le f_k(x)\le f(x)$ for every $k\ge K$, so $|f_k(x)-f(x)|\le\tfrac{\varepsilon}{2}<\varepsilon$ for every $k\ge K$. Since $\varepsilon$ was arbitrary, $(f_k(x))_{k\in\mathbb{N}}$ converges to $f(x)$ in the sense of Limit of a Sequence of Real Numbers.

Finally, $f(x)$ is an upper bound of $\{f_k(x):k\in\mathbb{N}\}$ by claim 1. Let $t$ be a real number with $t<f(x)$ and apply the argument above with $\varepsilon=f(x)-t$, which is positive: it produces $K\in\mathbb{N}$ with $f(x)-\tfrac{\varepsilon}{2}\le f_K(x)$, and since $0<\varepsilon$ we have $t=f(x)-\varepsilon<f(x)-\tfrac{\varepsilon}{2}\le f_K(x)$, so $t$ is not an upper bound. Hence $f(x)$ is the least upper bound of $\{f_k(x):k\in\mathbb{N}\}$.